我可以在熊猫中执行动态cumsum行吗？

Question

如果我有以下数据帧,那么派生如下: df = pd.DataFrame(np.random.randint(0, 10, size=(10, 1)))

    0
0   0
1   2
2   8
3   1
4   0
5   0
6   7
7   0
8   2
9   2

是否有一种有效的方式cumsum具有限制的行,每次达到此限制,以启动新的cumsum.达到每个限制(但行数很多)后,将使用总库存创建一行.

下面我创建了一个执行此操作的函数示例,但它非常慢,尤其是当数据框变得非常大时.我不喜欢我的功能是循环,我正在寻找一种方法来使它更快(我猜一种没有循环的方式).

def foo(df, max_value):
    last_value = 0
    storage = []
    for index, row in df.iterrows():
        this_value = np.nansum([row[0], last_value])
        if this_value >= max_value:
            storage.append((index, this_value))
            this_value = 0
        last_value = this_value
    return storage

如果你像我这样朗读我的函数:foo(df, 5) 在上面的上下文中,它返回:

   0
2  10
6  8

Answer 1

循环无法避免,但可以使用numba's 并行化njit:

from numba import njit, prange

@njit
def dynamic_cumsum(seq, index, max_value):
    cumsum = []
    running = 0
    for i in prange(len(seq)):
        if running > max_value:
            cumsum.append([index[i], running])
            running = 0
        running += seq[i] 
    cumsum.append([index[-1], running])

    return cumsum

假设您的索引不是数字/单调增加,则此处需要索引.

%timeit foo(df, 5)
1.24 ms ± 41.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit dynamic_cumsum(df.iloc(axis=1)[0].values, df.index.values, 5)
77.2 µs ± 4.01 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

如果索引是Int64Index类型,您可以将其缩短为:

@njit
def dynamic_cumsum2(seq, max_value):
    cumsum = []
    running = 0
    for i in prange(len(seq)):
        if running > max_value:
            cumsum.append([i, running])
            running = 0
        running += seq[i] 
    cumsum.append([i, running])

    return cumsum

lst = dynamic_cumsum2(df.iloc(axis=1)[0].values, 5)
pd.DataFrame(lst, columns=['A', 'B']).set_index('A')

    B
A    
3  10
7   8
9   4

%timeit foo(df, 5)
1.23 ms ± 30.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit dynamic_cumsum2(df.iloc(axis=1)[0].values, 5)
71.4 µs ± 1.4 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

njit 功能表现

perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.randint(0, 10, size=(n, 1))),
    kernels=[
        lambda df: list(cumsum_limit_nb(df.iloc[:, 0].values, 5)),
        lambda df: dynamic_cumsum2(df.iloc[:, 0].values, 5)
    ],
    labels=['cumsum_limit_nb', 'dynamic_cumsum2'],
    n_range=[2**k for k in range(0, 17)],
    xlabel='N',
    logx=True,
    logy=True,
    equality_check=None # TODO - update when @jpp adds in the final `yield`
)

对数 - 对数图显示对于较大的输入,生成器函数更快:

可能的解释是,随着N增加,附加到增长列表的开销dynamic_cumsum2变得突出.虽然cumsum_limit_nb只是必须yield.

Answer 2

循环不一定是坏的.诀窍是确保它在低级对象上执行.在这种情况下,您可以使用Numba或Cython.例如,使用生成器numba.njit:

from numba import njit

@njit
def cumsum_limit(A, limit=5):
    count = 0
    for i in range(A.shape[0]):
        count += A[i]
        if count > limit:
            yield i, count
            count = 0

idx, vals = zip(*cumsum_limit(df[0].values))
res = pd.Series(vals, index=idx)

为了演示使用Numba进行JIT编译的性能优势:

import pandas as pd, numpy as np
from numba import njit

df = pd.DataFrame({0: [0, 2, 8, 1, 0, 0, 7, 0, 2, 2]})

@njit
def cumsum_limit_nb(A, limit=5):
    count = 0
    for i in range(A.shape[0]):
        count += A[i]
        if count > limit:
            yield i, count
            count = 0

def cumsum_limit(A, limit=5):
    count = 0
    for i in range(A.shape[0]):
        count += A[i]
        if count > limit:
            yield i, count
            count = 0

n = 10**4
df = pd.concat([df]*n, ignore_index=True)

%timeit list(cumsum_limit_nb(df[0].values))  # 4.19 ms ± 90.4 µs per loop
%timeit list(cumsum_limit(df[0].values))     # 58.3 ms ± 194 µs per loop